EXPLANATION OF IODINE IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 9 , 2015 Iodine is a chemical element with symbol I and atomic number 53. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Iodine including the following ground state electron configuration: 1s22s22p63s2p63d104s24p64d105s25px25py25pz1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of iodine (from (E1 to E3 ) are the following: E1 = 10.45 , E2 = 19.13, and E3 = 33 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 10.45 eV = -E(5px2) + E(px1) Here the E(5px2) represents the binding energy of 5px2, while the E(5px1) represents the binding energy of 5px1.The charges (-48e) of (1s22s22p63s23p63d104s2 4p64d105s2) screen the nuclear charge (+53e) and for a perfect screening we would have ζ = 5. Note that the 5px2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5px2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5px1 consists of one electron, we apply the Bohr formula to write E(5px1) = (-13.6057)ζ2/n2 Therefore E1 = 10.45 eV = -E(5px2) + E(5px1) = - (16.95)ζ + 4.1) / n2 Since n = 5 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 257.15 = 0 Then, solving for ζ we get ζ = 5 . Here we see that the five electrons of 5p make a sub- shell of a spherical symmetry and lead to a perfect screening with ζ = 5, because they repel the clouds of 5s and 4d from symmetrical positions. Of course the two electrons of opposite spin ( say the 5px2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple Coulomb repulsion. This situation indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. EXPLANATION OF E2 = 19.13 eV = -E(5py2) + E(py1) Here the E(5py2) represents the binding energy of 5py2, while the E(5py1) represents the binding energy of 5py1. As in the case of E1 the charges (-48e) of (1s22s22p63s23p63d104s24p64d105s2) screen the nuclear charge (+53e) and for a perfect screening we would have ζ = 5. Note that the 5py2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5py2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5py1 consists of one electron, we apply the Bohr formula to write E(5py1) = (-13.6057)ζ2/n2 Therefore E2 = 19.13 eV = -E(5py2) + E(5py1) = - (16.95)ζ + 4.1) / n2 Since n = 5 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 474.15 = 0 Then, solving for ζ we get ζ = 7.66 > 5 Here we see that in the absence of one electron the electrons of 5p break the symmetry and lead to the deformations of electron clouds. Thus ζ = 7.66 > 5. EXPLANATION OF E3 = 33 eV = -E(5x1) Here the E(5px1) represents the binding energy of the 5px1. As in the cases of E1 and E2 the charges (-48e) of (1s22s22p63s23p63d104s2 4p64d10 5s2) screen the nuclear charge (+53e) and for a perfect screening we would have the same ζ = 5. However as in the case of E2 the electrons of 5p break more the symmetry and lead to the deformations of electron clouds. Thus ζ > 5. Since 5px1 is the one electron of parallel spin we apply the Bohr formula to write E3 = -E(5px1) = 33 eV = - (-13.6057)ζ2/n2 Therefore E3 = 33 eV = (13.6057)ζ2 / 52 Then we get ζ = 7.79 > 5 . Here 7.79 > 7.66 > 5 means that after the ionizations the electrons of 5p break more the symmetry and lead to a great deformation of electron clouds. Category:Fundamental physics concepts